1/100
The error in its area is then 2 percent....
area= side^2 let the symbol # denote error in measurement #area/area= 2(#length/length) #area/area*100= 2(#length/length)*100 percent error in area= 2*percent error in length=2% 2 per cent
The volume ( V ) of a cylinder is calculated using the formula ( V = \pi r^2 h ), where ( r ) is the radius and ( h ) is the height. If there is a 1 percent error in either the diameter or the height, the resulting error in volume can be found using the formula for propagation of errors. A 1 percent error in the diameter (which affects the radius) leads to approximately a 2 percent error in volume, while a 1 percent error in height results in a 1 percent error in volume. Therefore, the maximum error in volume can be about 2 percent when considering the diameter measurement error.
Sometimes you will take the absolute value of the percent error because your estimated number could be less than the theoretical, meaning the calculation is negative. But an absolute value is always positive. A percent error can be left as a negative though, and this would be perfectly acceptable (or even preferred) depending on what you're doing.Answer:In the sciences, a negative percent error indicates a low result. If you have a 0% error, then your observed (lab) result was exactly the same as the theoretical result. A 5% error could mean that your observed result was a little high. A negative percent error is possible; if your observed results were lower than the expected, then you would have a negative percent error. A -5% error could mean that your results were a little low. Having a negative percent error isn't worse than positive percent error -- it could mean the same thing. If you were to have a choice in having a 20% error and a -5% error, the negative percent error is more accurate.
between 41 and 51 percent
Yes. In some cases it can be much more than five percent.
5%
The percent for 2 out of 5 is 0.4%
2 days out of 5 - is 40 percent
Percent error.
The difference between low percent error and high percent error is one is low and the other is high
The percentage error in the area of the square will be twice the percentage error in the length of the square. This is because the error in the length affects both the length and width of the square, resulting in a compounded effect on the area. Therefore, if there is a 1 percent error in the length, the percentage error in the area would be 2 percent.